functions (1 Viewer)

[kractus]

 

[synthesisFR]

is the answer neither?
idk what i did but i did it in a way that is very unorthodox but kinda makes sense

 

[synthesisFR]

i think wat u do is
since the function is even (the top one), the only way is if each of the two on the left are also even bc thats the only way that makes sense
f(-x) + g(-x)= f(x) + g(x), since each are even, their addition will be symmetrical about the y and also even
however if they were both odd f(-x) + g(-x)= -f(x) - g(x) which sum to give -(f(x) + g(x)) which in a similar manner would also be odd if u think abt it graphically as they are symmetrical about the origin
so for this question, f(x-1) and f(1-x) must both be even,
also we get that f(1-x) = f[-(x-1)] = f(x-1)
So u have 2[f(x-1)] = 2x^2 + 8
so if [f(x-1)] = x^2 + 4 (this is shifted right by one unit)
the original function would have been shifted to the left by one unit compared to f(x-1) which would shift the parabola left meaning it wouldnt be even anymore and hence it would be neither
watch this be completely wrong and hence i will drop to standad 1 maths by next term

 

[unofficiallyred12]

easiest way to do should be considering how f(x+1-1) + f(1-(x+1)) = f(x) + f(-x) = 2(x+1)^2+8

 

[gazzaboy]

The function itself actually doesn't exist so the question isn't really a valid one. For example, if you substitute x = 1, you get f(-1) + f(1) = 8, but if you substitute x=2, you get f(-1) + f(1) = 16. Both of these can't be true at the same time.

I think the question is trying to get you to answer 'neither' by asking you to 'see what happens' if it is even or odd:

• If f(x) is even, then you have f(x-1) = f(1-x), so 2f(x-1) = 2x^2 + 8. If you let u = x-1 and rearrange, then you get f(u) = 2(u+1)^2 + 8, and this is not even, so we have a contradiction.
• If f(x) is odd, then you get f(x-1) - f(x-1) = 2x^2 + 8, which can't be true, since f(x-1) - f(x-1) needs to equal to the 'zero function'

However, the question itself is problematic so I wouldn't worry too much if you can't solve it